Why Ramanujan summation is actually -3/32

Liuba
3 min readMar 1, 2021

Few weeks ago my roommate effusively started telling me that the sum of all natural numbers is –1/12, that is:

1+2+3+4+5+6+7+8 +… = –1/12

I was quite surprised with this statement, specially hearing it from an engineering graduate student.

I myself have a Math background. I studied infinite series back and forth at the beginning of my career, and know some possible leaks that can appear in a demonstration of series convergence. Hence, I already knew what the trick was here, and also I know for a fact that this polemical sum actually equals -3/32. What most surprised me is discovering that the Ramanujan summation is used in string theory and quantum mechanics. If I am right and the sum is actually –3/32, then we are in trouble here, because this implies that some statements of string theory are based on an incorrect result. Will this impact string theory field?

First, let me convince you that the sum is -3/32.

I will prove the following two series:

1–1+1–1+1–1+… = 1/2

1–2+3–4+5–6+… = 1/4

The two results will then be used to prove that 1+2+3+4+5+6+…=–3/32

Let’s take the first series and denote it by A:

A = 1–1+1–1+1–1+…

Now substract A from 1:

1–A = 1–(1–1+1–1+1–…) =1–1+1–1+1–1+1–…

Look familiar? You are right, we got A again! Therefore,

1–A = A

1=2A

A = 1/2

Now, let’s take the second series and denote it by B:

B=1–2+3–4+5–6+…

Then we have that

B–A = (1–2+3–4+5–6+…)–(1–1+1–1+1–…)

B–A = (1–2+3–4+5–6+…) –1+1–1+1–1+…

Rearranging a little bit we get:

B–A = (1–1)+(–2+1)+(3–1)+(–4+1)+(5–1)+(–6+1)+…

B–A = 0–1+2–3+4–5+6–7+…

B–A = –B

2B = A

But A = 1/2, therefore

2B = 1/2

B = 1/4

Finally, let’s analyze C = 1+2+3+4+5+6+7+8+9+…

C–3B = (1+2+3+4+5+6+7+8+9+…) — (3–6+9–12+15–18+…)

C–3B = (1+2+3+4+5+6+7+8+9+…) –3+6–9+12–15+18-…

We will rearrange as follows:

C–3B = (1+2–3) + (3+6) + (4+5–9) + (6+12) + (7+8–15) +(9+18)+ …

where last term comes from the second summand 3B and the first one or two terms come from C.

When realizing the sum inside the parentheses, we get the following:

C–3B= 0 + 9 + 0 + 18+0+27+0+… = 9+18+27+…

C–3B = 9(1+2+3+…) = 9C

B = –8C

But B = 3/4. So, substituting:

3/4 = –8C

C = –3/32

We have proven that

1+2+3+4+5+6+7+8+9+… = –3/32

Mind blowing, isn’t it?

Obviously, as you might have guessed, this summation has also been proven to be -1/12. Is there a particular reason of why we should choose -1/12 rather than -3/32? What about any other number (yes, by rearranging the series we can actually get to any number we want!).

The main concern I have is in how would string theory change if we use my proof instead of Ramanujan summation? I know nothing about string theory, but I am really curious of possible effects that might occur. Let me know your opinion about this. I would be happy to have a discussion and learn of what you think about this topic!

Best,

--

--

Liuba

Ph.D. Computer Science student at Rice University. Interests: human robot interaction, autonomous driving, human behavior. Please, contact me for any questions!